Calculating effective mass when Spring itself is massive in Mechanics problems.

So, this is a post regarding a general result which is of a lot of help if remembered in Mechanics problems. The derivation is of how to calculative effective mass of block and spring attached if spring is massive.

Let's consider the mass of spring to be mm.

So let's begin,

So let us take a small element of mass dmdm of the spring, so total energy can be calculated by adding all the length elements' kinetic energy, and requires the following integral:

E=12u2dmE = \int \dfrac{1}{2}u^{2} dm

Since spring is uniform,

dm=(dyL)mdm = (\dfrac{dy}{L})m

E=0L12u2(dyL)mE = \int_{0}^{L} \dfrac{1}{2}u^{2}(\dfrac{dy}{L})m

The velocity of each mass element of the spring is directly proportional to its length, i.e.

u=(vyL)u = (\dfrac {vy}{L})

E=12mL0L(vyL)2dyE = \dfrac{1}{2} \dfrac{m}{L} \int_{0}^{L} (\dfrac{vy}{L})^{2} dy

E=12m3v2E= \dfrac{1}{2} \dfrac{m}{3} v^{2}

Comparing to the expected original kinetic energy formula,

K.E.=12mv2K.E. = \dfrac{1}{2}mv^{2}

we can conclude that effective mass of spring in this case is m3\dfrac{m}{3}

The EFFECTIVE MASS of the spring in a spring-mass system when using an ideal spring of uniform linear density is 13\dfrac{1}{3} of the mass of the spring.

And we're done.

#Physics #Mechanics #Springs #EffectiveMass

Note by Ritvik Choudhary
6 years, 4 months ago

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1 vote

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Comments

how we came to know that velocity of each mass element of the spring is directly proportional to its length????

Nikhil Tanwar - 6 years, 4 months ago

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It's true that this result is based only on the assumption that velocity of each mass element of the spring is directly proportional to its length, but since above is technique used for solving certain problems in classical mechanics where you can generally take the above assumption as valid.

Ritvik Choudhary - 6 years, 3 months ago

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Thanks Ritvik

Nikhil Tanwar - 6 years, 3 months ago

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@Nikhil Tanwar Glad to help :)

Ritvik Choudhary - 6 years, 3 months ago

This is only True when we assume Liniar relation of velocity of massive spring ! So it is not the genral result !

Karan Shekhawat - 6 years, 3 months ago

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how can we assume it..?? can you pls explain how is velocity linearly proportional to distance of point from the fixed end..

Shara Handrale - 3 years, 3 months ago

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It is only true when the spring is stretching/compressing uniformly,i.e. when the end point of the spring is stretched by x, then the mid-point is stretched by (x/2). So the element of the spring which is at y distance from the origin should stretch by (xy/L) , where L is the length of the spring. So from this we can see that the velocity of that element is (vy/L) where v is the velocity of the end point of the spring. Hope this helps..

Nikhil Tanwar - 3 years, 1 month ago
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